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I have a problem and a proposed solution. I want to know if I have done it correctly.

Problem Statement: Let $V=F^n$ be the space of column vectors. Prove that every subspace $W$ of $V$ is the space of solutions of some system of homogeneous linear equations $AX=0$.

My solution: The null space of an $m$ x $n$ matrix $A$, written as $Nul A$, is the set of all solutions to the homogeneous equation $Ax = 0$. The null space of an $m$ x $n$ matrix $A$ is a subspace of $R^n$. Equivalently, the set of all solutions to a system $Ax = 0$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $R^n$.

Then I proceeded to prove that the properties of a subspace hold for the null space, and hence the null space is a subspace of $R^n$. Please tell me if this is right... this problem has been on my head all day!!

Thanks!

  • I'm afraid it is not correct. You have proven that the set of all solutions of $Ax=0$ is a subspace, but what was asked is an opposite direction, i.e., that for each subspace $S$ there is a matrix $A$ such that $S$ is equal to the set of solutions of $Ax = 0$. – Vedran Šego Sep 23 '13 at 00:37
  • Ok thanks for the heads up. I am really stuck –  Sep 23 '13 at 00:42
  • Do you know what the dual space is? – Pedro Sep 23 '13 at 00:45
  • @Peter Tamaroff No, I'm afraid not. –  Sep 23 '13 at 00:46
  • @Anonymous Hm, maybe one can do something without mentioning it explicitly. – Pedro Sep 23 '13 at 00:47
  • Do you know about the orthogonal projection of a vector onto a subspace? – Michael Albanese Sep 23 '13 at 01:05
  • @MichaelAlbanese A little bit, but not enough to put it into a proof format. You see, I'm quite rusty on my linear algebra. I am taking a general algebra course with some unwanted linear algebra thrown in –  Sep 23 '13 at 01:09
  • Let's see if you know enough. Suppose $W$ is a subspace of $F^n$. If the orthogonal projection of a vector $v \in F^n$ onto $W$ is itself, what can we say about $v$? – Michael Albanese Sep 23 '13 at 01:12
  • Is my proof below on the right track? –  Sep 23 '13 at 01:18
  • It is on the right track but you need to explain why $M = W$. – Michael Albanese Sep 23 '13 at 01:32

4 Answers4

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You have proved that the solution set of the system $Ax = 0$ is a subspace of $F^n$. What the question asks you to do is show that any subspace is the solution set of some system of equations. That is, you have to show that, given a subspace $W$ of $F^n$, there is a matrix $A$ such that $\{x \in F^n \mid Ax = 0\} = \operatorname{Nul}A = W$.

Michael Albanese
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Disclaimer: Not entirely sure about this.

Proof: Let $V = \mathbb{F^{n}}$. Let $W$ be a subspace of $V$. By definition, every subspace of $V = \mathbb{F}^n$ must contain the $0$ vector and must be closed under addition and scalar multiplication. Therefore, all linear combinations of the system $Ax = a_1x_1+\cdots+a_nx_n = 0$ are contained in $W$, so it is clear that $AX = 0 \in W$, and as a result, $W$ can be represented as the solution space to the homogenous equation. $\Box$

St Vincent
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    I realize this question is very old, but I stumbled on it in the review queues, and I can't for the life of me figure out how this answers the question. Or what it's trying to say at all. What are $A$ and $x$? The question is asking how to find an $A$ such that the solutions to the equation $AX=0$ are precisely the elements of $W$. – jgon Mar 12 '19 at 02:03
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Let $W$ be your subspace. If $W$ is all of the space, it is the kernel of the null matrix. If it is $\{0\}$, it is the kernel of any nonsingular matrix. So assume your subspace is proper, and let $\{w_1,\ldots,w_s\}$ be a basis. Extend this to a basis $\{w_1,\ldots,w_s,w_{s+1},\ldots,w_n\}$ of your space, let $\{w_{s+1},\ldots,w_n\}=W'$. Define the transformation $f:V\to V$ as the projection onto $W'$ through $W$. Then $W=\ker f$, and you can consider the matrix of $f$ in the canonical basis to get an appropriate system of equations.

Alternatively, use the canonical inner product: if $W$ is your subspace, consider $W^\perp$. Then use that $W=(W^\perp)^{\perp}$ to get a set of equations for $W$, namely of the form $w_i\cdot x=0$ for $i=1,2,\ldots, r$ where $\{w_1,\ldots,w_r\}$ is a base for $W^{\perp}$.

Pedro
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Disclaimer: Not entirely sure about this.

Here is a totally different solution that I have thought up out of desperation, but I'm not sure about it.

Let $W$ be a proper subspace of $V$. Then, as $\dim(W) + \dim(W^\perp) = \dim(V)$, we have $W^\perp$ not equal to $0$. Construct the matrix $A^t$ by taking its columns to be a basis for $W^\perp$. Let $M=\{x \in V : Ax = 0\}$ be the solution set under consideration. Then $W=M$. Next assume that $W=V$. Taking $A$ to equal the $0$ matrix, we are done.

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    If you are going to use braces (i.e. { and }) in math mode, you need to add a backslash beforehand (i.e. \{ and \}), otherwise they don't appear. The same goes for special symbols LaTeX uses for other purposes (such as $$$, $%$, etc.). – Michael Albanese Sep 23 '13 at 01:34
  • I think you mean $W$ is a proper subspace, not subset. – Michael Albanese Sep 23 '13 at 01:38
  • Yeah, you're right. I fixed it. Thanks for all your help! –  Sep 23 '13 at 01:43